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Trihybrid Cross Calculator

Calculate trihybrid cross outcomes for three independent gene pairs. Shows genotype and phenotype probabilities for all eight phenotype classes.

Frequently Asked Questions

What is the phenotype ratio for a standard trihybrid cross?

Crossing AaBbCc × AaBbCc produces an expected 27:9:9:9:3:3:3:1 phenotype ratio across eight classes when all three genes assort independently and each has simple dominant-recessive inheritance.

How is the trihybrid ratio derived?

Independent assortment allows each gene pair to be treated separately. Each Aa × Aa cross gives a 3:1 dominant-to-recessive ratio. Multiplying across three genes: (3+1)^3 = 64 total combinations, which expands to 27:9:9:9:3:3:3:1.

What is the probability of the triple homozygous dominant (AABBCC) offspring?

From AaBbCc × AaBbCc, the probability of AA at each locus is 1/4. For three independent loci: 1/4 × 1/4 × 1/4 = 1/64, or about 1.56% of offspring.

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Estimates for informational purposes only.

Important Disclaimer: Estimates for informational purposes only.

This calculator provides estimates for informational purposes only. Results are based on assumptions and may not reflect actual outcomes. Consult qualified professionals in relevant fields before making important decisions based on these results.

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How This Calculator Works

A trihybrid cross involves three independent gene pairs simultaneously. This calculator determines the probability of each phenotype class in the offspring of two parents, where each parent can be homozygous dominant (AA), heterozygous (Aa), or homozygous recessive (aa) at each of the three loci.

Because the three gene pairs are assumed to assort independently (Mendel's Law of Independent Assortment), the probability of any combined phenotype is simply the product of the individual gene probabilities. This multiplicative approach is far more practical than drawing a massive 8x8 Punnett grid, which would contain 64 cells.

Formula: Independent Assortment

For each gene pair, the calculator first determines the probability of the dominant phenotype and the recessive phenotype from the two parental genotypes. Then it multiplies across all three genes:

P(A_B_C_) = P(A_) x P(B_) x P(C_)
P(A_B_cc) = P(A_) x P(B_) x P(cc)
P(A_bbC_) = P(A_) x P(bb) x P(C_)
And so on for all eight phenotype classes

The eight probabilities always sum to 1 (100%), covering every possible phenotype combination.

Standard Ratios: AaBbCc x AaBbCc

The most commonly tested case in genetics courses is the triple heterozygote cross AaBbCc x AaBbCc. For each gene, one Aa x Aa cross gives P(dominant) = 3/4 and P(recessive) = 1/4. Multiplying across three genes produces the famous 27:9:9:9:3:3:3:1 ratio across 64 total offspring units:

27/64 - dominant at all three loci (A_B_C_)
9/64 each - dominant at two loci, recessive at one (three such classes)
3/64 each - dominant at one locus, recessive at two (three such classes)
1/64 - recessive at all three loci (aabbcc)

This ratio assumes complete dominance at all three loci and independent assortment. Linked genes or incomplete dominance will change the actual proportions observed.

Worked Example

Suppose you cross AABbcc x AaBBCc. Work gene by gene:

Gene A: AA x Aa gives P(A_) = 1, P(aa) = 0
Gene B: Bb x BB gives P(B_) = 1, P(bb) = 0
Gene C: cc x Cc gives P(C_) = 1/2, P(cc) = 1/2

So P(A_B_C_) = 1 x 1 x 1/2 = 1/2, and P(A_B_cc) = 1 x 1 x 1/2 = 1/2. All other phenotype classes have probability zero because gene A and gene B are fully dominant in all offspring from these parent combinations. The calculator handles any combination of parental genotypes automatically.

Tips and FAQ

What does A_ mean?

The underscore is a wildcard representing either allele - A or a. A_ means the offspring has at least one dominant A allele, so it expresses the dominant phenotype for gene A regardless of whether it is AA or Aa.

Why does the AaBbCc probability sometimes appear as zero?

The probability shown is specifically for the genotype AaBbCc (heterozygous at all three loci). If either parent is homozygous (AA or aa) for any gene, it is impossible for offspring to be heterozygous at that locus, so the triple-heterozygote probability drops to zero.

Does this work for linked genes?

No. This calculator assumes independent assortment, which holds when the three genes are on different chromosomes or are far apart on the same chromosome. For linked genes, recombination frequencies must be factored in and a simple product rule does not apply.

What about more than three genes?

The same multiplication principle extends to any number of independently assorting genes. For n genes, there are 2^n phenotype classes. For four genes, that is 16 classes with a total of 256 units in the standard quadrihybrid ratio.